Graph Isomorphism and Related Problems Graph Isomorphism Graph Isomorphism Two graphs G1 = (V1; E1) and G2 = (V2; E2) are isomorphic if there is a bijection ': V1 ! V2 such that, for all u; v 2 V1, uv 2 E1 if and only if '(u)'(v) 2 E2. Isomorphism of two graphs G1 and G2 is denoted as G1 ' G2. 2 / 10 Graph Isomorphism { Related Problems Subgraph Isomorphism I Determine if G is isomorphic to a subgraph of H. Largest Common Subgraph I For two given graphs G1 and G2, find the largest graph H such that H is isomorphic to a subgraph of G1 and to a subgraph of G2. Theorem There is (probably) no polynomial time algorithm to solve the Subgraph Isomorphism problem. c In 2015, Babai presented a proof that there is a O 2log n time algorithm to solve the Graph Isomorphism problem. (Result is not peer reviewed yet.) 3 / 10 Isomorphism of Trees Isomorphism of Trees Centers I A vertex v is center of G, if ecc(v) = rad(G). I A tree has at most two centers which can be found in linear time. In case of two, both are adjacent. Finding ceters in trees I Pick an arbitrary vertex v. I Find a vertex x with d(v; x) = ecc(v). I Find a vertex y with d(x; y) = ecc(x). I Then, ecc(x) = ecc(y) = diam(T) and the vertices in the middle of the path from x to y are centers of T (two vertices if d(x; y) is odd, one vertex if d(x; y) is even). Note: There is a linear time algorithm to compute the eccentricity of every vertex in a tree in linear time. 5 / 10 Isomorphism of Trees Rooted Tree We say (V; E; r) is a tree with root r if (V; E) is a tree and r 2 V. Isomorphism of Rooted Trees Two rooted trees T1 = (V1; E1; r1) and T2 = (V2; E2; r2) are isomorphic if there is a bijection ': V1 ! V2 such that, for all u; v 2 V1, uv 2 E1 if and only if '(u)'(v) 2 E2 and '(r1) = '(r2). T1 T2 T3 T1 and T3 are isometric. However, T2 is not isometric to T1 and T3. 6 / 10 Isomorphism of Trees Theorem If there is an O(f (n)) time algorithm A to check if two rooted trees are isomorphic, then there is an O(f (n)) time algorithm A∗ to check if two ordinary trees are isomorphic. ∗ Proof: We design A (T1; T2) as follows: Find the centers of T1 and T2. Then, there are three cases. C1 Both have only one center (c1 and c2). Retrun A(T1; c1; T2; c2) 0 0 C2 Both have two centers (c1, c1, c2, and c2). 0 Retrun A(T1; c1; T2; c2) _A(T1; c1; T2; c2) C3 Different number of centers. Return False Therefore, it is enough if we can check rooted trees. 7 / 10 Aho-Hopcroft-Ullman-Algorithm Let T1 and T2 be two rooted trees with corresponding roots r1 and r2. With Fi;j, we denote the forest created by removing all vertices v 2 Ti with d(v; ri) ≤ j. Theorem T1 is isometric to T2 if and only if, for all i, F1;i is isometric to F2;i. Algorithm idea: I Check bottom-up if F1;i is isometric to F2;i. I Note: The lowest layer contains only leaves. I Use that F1;i+1 is isometric to F2;i+1 when checking F1;i and F2;i. 8 / 10 Aho-Hopcroft-Ullman-Algorithm Input: Two rooted trees T1 and T2 with the corresponding roots r1 and r2 and with height h. 1 Compute the depth of each vertex in T1 and T2. 2 To each leaf, assign the integer 0. 3 For i = h − 1 DownTo 1 4 hAssign integers to all nodes with depth i.i 5 T1 and T2 are isomorphic if and only if r1 and r2 have the same integer assigned. 9 / 10 Aho-Hopcroft-Ullman-Algorithm 1 Procedure AssignIntegers(i) 2 For Each v 2 Vi (the set of non-leaf vertices with depth i) 3 Create a sorted tuple tv containing the integers assigned to the children of v. 4 Let S1 and S2 be the sets of tuples created for vertices in T1 and T2 with depth i. 5 Sort S1 and S2 lexicographically and compare them. If S1 6= S2, Stop. T1 and T2 are not isomorphic. 6 Assign integers, continuously and beginning with 1, to vertices in Vi such that two vertices have the same integer assigned if and only if they have equal tuples. 10 / 10.